Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 630

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\[1)\ 2\sin^{2}x = 1 + \frac{1}{3}\sin{4x}\]

\[1 - \cos{2x} = 1 + \frac{2}{3}\sin{2x} \bullet \cos{2x}\]

\[\frac{2}{3}\sin{2x} \bullet \cos{2x} + \cos{2x} = 0\]

\[\cos{2x} \bullet \left( \frac{2}{3}\sin{2x} + 1 \right) = 0\]

\[\cos{2x} = 0\]

\[2x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[\frac{2}{3}\sin{2x} + 1 = 0\]

\[\frac{2}{3}\sin{2x} = - 1\]

\[\sin{2x} = - \frac{3}{2}\]

\[корней\ нет.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[2)\ 2\cos^{2}{2x} - 1 = \sin{4x}\]

\[1 + \cos{4x} - 1 - \sin{4x} = 0\]

\[\cos{4x} - \sin{4x} = 0\ \ \ \ \ |\ :\cos{4x}\]

\[1 - tg\ 4x = 0\]

\[tg\ 4x = 1\]

\[4x = arctg\ 1 + \pi n\]

\[4x = \frac{\pi}{4} + \pi n\]

\[x = \frac{1}{4} \bullet \left( \frac{\pi}{4} + \pi n \right) = \frac{\pi}{16} + \frac{\text{πn}}{4}.\]

\[Ответ:\ \ \frac{\pi}{16} + \frac{\text{πn}}{4}.\]

\[3)\ 2\cos^{2}{2x} + 3\cos^{2}x = 2\]

\[2\cos^{2}{2x} + \frac{3}{2}\left( 1 + \cos{2x} \right) = 2\]

\[2\cos^{2}{2x} + 1,5 + 1,5\cos{2x} - 2 = 0\]

\[2\cos^{2}{2x} + 1,5\cos{2x} - 0,5 = 0\]

\[Пусть\ y = \cos{2x},\ тогда:\]

\[2y^{2} + 1,5y - 0,5 = 0\]

\[4y^{2} + 3y - 1 = 0\]

\[D = 9 + 16 = 25\]

\[y_{1} = \frac{- 3 - 5}{2 \bullet 4} = - 1;\]

\[y_{2} = \frac{- 3 + 5}{2 \bullet 4} = \frac{1}{4}.\]

\[\cos{2x} = - 1\]

\[2x = \pi - \arccos 1 + 2\pi n\]

\[2x = \pi + 2\pi n\]

\[x = \frac{1}{2} \bullet (\pi + \pi n)\]

\[x = \frac{\pi}{2} + \pi n.\]

\[\cos{2x} = \frac{1}{4}\]

\[2x = \pm \arccos\frac{1}{4} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \arccos\frac{1}{4} + 2\pi n \right)\]

\[x = \pm \frac{1}{2}\arccos\frac{1}{4} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ \]

\[\pm \frac{1}{2}\arccos\frac{1}{4} + \pi n.\]

\[4)\ \left( \sin x + \cos x \right)^{2} = 1 + \cos x\]

\[\cos^{2}x + \sin^{2}x + 2\sin x \bullet \cos x =\]

\[= 1 + \cos x\]

\[1 + 2\sin x \bullet \cos x - 1 - \cos x = 0\]

\[2\sin x \bullet \cos x - \cos x = 0\]

\[\cos x \bullet \left( 2\sin x - 1 \right) = 0\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n\]

\[x = \frac{\pi}{2} + \pi n.\]

\[2\sin x - 1 = 0\]

\[2\sin x = 1\]

\[\sin x = \frac{1}{2}\]

\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ \]

\[\ ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]